What is the time complexity of Strong Password Checker?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Strong Password Checker?

This problem uses String, Greedy, Heap (Priority Queue). Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Strong Password Checker?

Insight 1: The problem can be broken down into independent subproblems: ensuring sufficient length, fulfilling character type requirements, and handling repeating characters. Insight 2: When the password is too short, insertions are the primary focus. When too long, deletions take precedence. When the length is within bounds, replacements are the main concern for repeating characters.

Which companies ask LeetCode 420. Strong Password Checker in interviews?

This problem is frequently asked at Devtron, Flipkart, Wix.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Strong Password Checker - LeetCode 420 Solution

Strong Password Checker - Complete Solution Guide

Strong Password Checker is LeetCode problem 420, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A password is considered strong if the below conditions are all met: It has at least 6 characters and at most 20 characters. It contains at least one lowercase letter, at least one uppercase letter, and at least one digit . It does not contain three repeating characters in a row (i.e., "B aaa bb0" is weak, but "B aa b a 0" is strong). Given a string password , return the minimum number of steps required to make password strong. if password is already strong, return 0 . In one step, you can: Inse

Detailed Explanation

The problem requires you to determine the minimum number of operations (insertions, deletions, or replacements) needed to transform a given string `password` into a strong password. A strong password must satisfy three conditions: 1) length between 6 and 20 characters, 2) contain at least one lowercase letter, one uppercase letter, and one digit, and 3) not have three consecutive repeating characters. The goal is to find the fewest steps to satisfy all conditions. If the password already meets these criteria, return 0.

Solution Approach

The provided solution employs a greedy approach to minimize the number of operations. It first checks for the presence of lowercase, uppercase, and digit characters and counts the number of missing character types. Then, it identifies repeating character sequences of length 3 or more. Based on the password's length, it applies different strategies: if the password is too short, it prioritizes insertions to reach a length of 6, taking into account any missing character types. If the password is too long, it prioritizes deletions, strategically removing characters from repeating sequences to minimize the number of replacements needed later. If the length is within the acceptable range, it calculates the number of replacements required to eliminate repeating sequences and ensures the password meets the missing character type requirement. The solution efficiently combines insertions, deletions, and replacements to achieve a strong password with the fewest possible operations.

Step-by-Step Algorithm

Step 1: Determine the length of the password and initialize counters for lowercase, uppercase, and digit character types.
Step 2: Check if the password contains at least one lowercase letter, one uppercase letter, and one digit. Count the number of missing character types.
Step 3: Identify all repeating sequences of characters with length >= 3. Store the lengths of these sequences in the `repeats` array/list.
Step 4: If the password length is less than 6, calculate the minimum number of operations needed as the maximum of (6 - password length) and the number of missing character types.
Step 5: If the password length is between 6 and 20 (inclusive), calculate the number of replacements required to fix repeating sequences (r // 3 for each repeating sequence of length r). The minimum number of operations is the maximum of the number of missing character types and the required replacements.
Step 6: If the password length is greater than 20, calculate the number of deletions needed to bring the length down to 20. Prioritize deletions from repeating sequences to minimize replacements. Deletions are processed from repeating sequences where the length mod 3 is 0 first, then sequences where the length mod 3 is 1, and finally, remaining deletions are applied to any sequences. Update the number of necessary replacements accordingly.
Step 7: Return the total number of operations, which is the sum of deletions (if any) and the maximum of missing character types and any remaining replacements.

Key Insights

Insight 1: The problem can be broken down into independent subproblems: ensuring sufficient length, fulfilling character type requirements, and handling repeating characters.
Insight 2: When the password is too short, insertions are the primary focus. When too long, deletions take precedence. When the length is within bounds, replacements are the main concern for repeating characters.
Insight 3: The repeats are handled greedily starting with mod 0 because they have the maximum number of replaceable characters per deletion. then mod 1 for similar reasons

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Greedy, Heap (Priority Queue).

Companies

Asked at: Devtron, Flipkart, Wix.

Strong Password Checker | LeetCode 420

Hard

String Greedy Heap (Priority Queue)

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int strongPasswordChecker(String password) {
        int n = password.length();

        boolean has_lower = false;
        for (char c : password.toCharArray()) {
            if ('a' <= c && c <= 'z') {
                has_lower = true;
                break;
            }
        }

        boolean has_upper = false;
        for (char c : password.toCharArray()) {
            if ('A' <= c && c <= 'Z') {
                has_upper = true;
                break;
            }
        }

        boolean has_digit = false;
        for (char c : password.toCharArray()) {
            if ('0' <= c && c <= '9') {
                has_digit = true;
                break;
            }
        }

        int missing_types = 3 - ((has_lower ? 1 : 0) + (has_upper ? 1 : 0) + (has_digit ? 1 : 0));

        java.util.List<Integer> repeats = new java.util.ArrayList<>();
        int i = 0;
        while (i < n) {
            int j = i;
            while (j < n && password.charAt(j) == password.charAt(i)) {
                j++;
            }
            int length = j - i;
            if (length >= 3) {
                repeats.add(length);
            }
            i = j;
        }

        if (n < 6) {
            return Math.max(6 - n, missing_types);
        } else if (n <= 20) {
            int replacements_for_repeats = 0;
            for (int r : repeats) {
                replacements_for_repeats += r / 3;
            }
            return Math.max(missing_types, replacements_for_repeats);
        } else {
            int deletes_to_do = n - 20;

            int replacements_for_repeats = 0;
            int[] mods = new int[]{0, 0, 0};
            for (int r : repeats) {
                replacements_for_repeats += r / 3;
                mods[r % 3]++;
            }

            int use_on_mod0 = Math.min(deletes_to_do, mods[0]);
            deletes_to_do -= use_on_mod0;
            replacements_for_repeats -= use_on_mod0;

            int use_on_mod1 = Math.min(deletes_to_do / 2, mods[1]);
            deletes_to_do -= use_on_mod1 * 2;
            replacements_for_repeats -= use_on_mod1;

            int use_on_any = deletes_to_do / 3;
            replacements_for_repeats -= use_on_any;

            return (n - 20) + Math.max(missing_types, replacements_for_repeats);
        }
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 420: Strong Password Checker - Detailed explanation and solution approaches.

Topics

String Greedy Heap (Priority Queue)

Companies

DevtronFlipkartWix

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:420

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Determine the length of the password and initialize counters for lowercase, uppercase, and digit character types.
2
Step 2: Check if the password contains at least one lowercase letter, one uppercase letter, and one digit. Count the number of missing character types.
3
Step 3: Identify all repeating sequences of characters with length >= 3. Store the lengths of these sequences in the `repeats` array/list.
4
Step 4: If the password length is less than 6, calculate the minimum number of operations needed as the maximum of (6 - password length) and the number of missing character types.
5
Step 5: If the password length is between 6 and 20 (inclusive), calculate the number of replacements required to fix repeating sequences (r // 3 for each repeating sequence of length r). The minimum number of operations is the maximum of the number of missing character types and the required replacements.
6
Step 6: If the password length is greater than 20, calculate the number of deletions needed to bring the length down to 20. Prioritize deletions from repeating sequences to minimize replacements. Deletions are processed from repeating sequences where the length mod 3 is 0 first, then sequences where the length mod 3 is 1, and finally, remaining deletions are applied to any sequences. Update the number of necessary replacements accordingly.
7
Step 7: Return the total number of operations, which is the sum of deletions (if any) and the maximum of missing character types and any remaining replacements.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operations are iterating through the password string to find repeating sequences and character type occurrences. These operations are linear in the length of the password. The other operations such as counting missing characters or performing calculations based on the repeating sequences are done in linear time with respect to the number of repeating sequences which is bounded by O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

A dynamic programming approach could potentially be used to solve this problem, but given the problem's constraints, it would likely be less efficient than the greedy approach. DP might involve considering all possible combinations of insertions, deletions, and replacements, which can lead to a much higher time complexity.

Time: O(n^3) or higher, depending on the DP formulation.Space: O(n^2) or higher.

Trade-offs: DP is generally not recommended for this problem because the greedy approach is simpler and more efficient for the given constraints. DP could theoretically handle more complex constraints or scoring functions, but that's not required here.

✨

Key Insights

1Insight 1: The problem can be broken down into independent subproblems: ensuring sufficient length, fulfilling character type requirements, and handling repeating characters.
2Insight 2: When the password is too short, insertions are the primary focus. When too long, deletions take precedence. When the length is within bounds, replacements are the main concern for repeating characters.
3Insight 3: The repeats are handled greedily starting with mod 0 because they have the maximum number of replaceable characters per deletion. then mod 1 for similar reasons

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty password (or password of length 1). The algorithm correctly calculates the number of insertions required to reach a length of 6 and satisfy any missing character type requirements.
Edge case 2: Password with a large number of repeating characters (e.g., 'aaaaaaaaaaaaaaaa'). The algorithm efficiently identifies these sequences and calculates the minimum number of replacements and/or deletions required.
Edge case 3: Passwords between length 6 and 20 that already fulfill type and repeating character criteria. The algorithm should return 0.
Edge case 4: password with no repeating characters, but is not of sufficient length or lacking types: The algorithm should only return the amount of missing characters/length, as the repeats section will not impact this at all.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the interaction between insertions, deletions, and replacements, particularly when the password is too long or too short. It's important to prioritize the right operations at each stage to minimize the overall number of steps.

✗Common mistake 2: Failing to account for all three character type requirements (lowercase, uppercase, and digit) when calculating the number of missing types.

✗Common mistake 3: Not efficiently identifying and grouping repeating character sequences, leading to an inefficient calculation of replacement operations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Password strength validation systems. This algorithm can be used as a component of password strength checkers to provide users with feedback on how to improve their passwords and meet security requirements.

🏦Real-world scenario 2: Account security policies. Organizations can use this algorithm to enforce password policies that ensure a minimum level of password strength across user accounts, reducing the risk of security breaches.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Strong Password Checker"?

For this hard-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?